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BaLLatris [955]
3 years ago
13

To determine the concentration of an EDTA solution, 4.11 g of Zn metal was used. The volume of EDTA solution needed to reach the

endpoint was 28.26 mL. What was the concentration (in molarity) of the EDTA solution?
Chemistry
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

2.23M

Explanation:

Molarity of a solution is calculated thus

Molarity = number of moles (n) ÷ volume (V)

According to this question, 4.11g of Zn metal was used in order to reach a volume of EDTA solution of 28.26 mL.

28.26mL = 28.26/1000

= 0.02826L

Using mole = mass/molar mass to calculate no. of moles of Zn

Mole = 4.11/65.4

mole = 0.0628mol

Molarity = 0.0628 ÷ 0.02826

Molarity = 2.23M

The concentration of the EDTA solution used is 2.23M

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Answer

Mg2Cl2

Mg + mg = mg2

cl + cl = cl2

Mg2 + cl2 =mg2cl2

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A gram is one million times larger than a microgram but one billion times smaller than a _______________.
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Answer : A hectogram

Explanation:
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n the 1990’s, the Boeing Co. developed a process for treating wastewater from electroplating and printed circuit board manufactu
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Final concentrations:

Cu²⁺ = 0

Al³⁺ = 3.13 mmol/L = 84.51 mg/L

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Explanation:

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Al: 27 g/mol ∴ 100 mg = 3.7 mmol

Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol

3 mol Cu²⁺ _______ 2 mol Al

4.7 mmol Cu²⁺ _____ x

x = 3.13 mmol Al

4.7 mmol of Cu²⁺ will be consumed.

3.13 mmol of Al will be consumed.

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2 years ago
How many formula units are contained in 2.35mol of Iron (III) Oxide
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3 years ago
: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne
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Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

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